Author Topic: TUT0302  (Read 264 times)


  • Jr. Member
  • **
  • Posts: 6
  • Karma: 0
    • View Profile
« on: October 04, 2019, 02:01:09 PM »
Problem: Show that the given equation becomes exact when multiplied by the integrating factor and solve the equation.

$My=x^{2}3y^{2} \neq Nx=\frac{1}{2}x^{2}$
Therefore it is not exact.

Multiply $\mu(\frac{1}{xy^{3}})$ to both side:
$My=0 = Nx=0$
Now it is exact.

Integrate M with respect to $x$ we get:
Take derivative with respect to $y$ on both side:
$\phi y=N=0+h'(y)$
So $h'(y)=(\frac{1}{y^{3}}+\frac{1}{y})$
$h(y)=\int y^{-3}+y^{-1}dy$

General Solution: $\frac{1}{2}x^{2}-\frac{1}{4}y^{-4}+lny=C$